$f(x, y) = (x^2 + y^2) \sqrt{1 - \dfrac{y^2}{x^2 + y^2}}$ We have a change of variables: $\begin{aligned} x &= X_1(r, \theta) = r \cos(\theta) \\ \\ y &= X_2(r, \theta) = r \sin(\theta) \end{aligned}$ What is $f(x, y)$ under the change of variables? Choose 1 answer: Choose 1 answer: (Choice A) A $r^2 |\cos(\theta)|$ (Choice B) B $r^3 |\sin(\theta)|$ (Choice C) C $r^3 |\cos(\theta)|$ (Choice D) D $r^2 |\sin(\theta)|$
Explanation: When applying a change of variables, we substitute the new definition for $x$ and $y$ into the original equation. The original equation: $f(x, y) = (x^2 + y^2) \sqrt{1 - \dfrac{y^2}{x^2 + y^2}}$ Let's substitute $X_1(r, \theta)$ for $x$ and $X_2(r, \theta)$ for $y$. We can rewrite every $x^2 + y^2$ using the trigonometric identity that $\cos^2(\theta) + \sin^2(\theta) = 1$ : $\begin{aligned} x^2 + y^2 &= r^2\cos^2(\theta) + r^2\sin^2(\theta) \\ \\ &= r^2(\cos^2(\theta) + \sin^2(\theta)) \\ \\ &= r^2 \end{aligned}$ Therefore: $\begin{aligned} f(x, y) &= r^2 \sqrt{1 - \dfrac{r^2\sin^2(\theta)}{r^2}} \\ \\ &= r^2 \sqrt{1 - \sin^2(\theta)} \\ \\ &= r^2 \sqrt{\cos^2(\theta)} \\ \\ &= r^2 |\cos(\theta)| \end{aligned}$ Therefore, under the change of variables, $f(x, y)$ becomes: $r^2 |\cos(\theta)|$